Understanding differentation formulae

Why is \(\frac{d}{dx} \left(x^{n} \right) = nx^{n-1}\) ?

We can thus think of the gradient between \(x\) and \(a\) being equal to the change in the y axis divided by the change in the x axis, i.e.

\[ \frac{\Delta y}{\Delta x} = \frac{f(x)-f(a)}{x-a} = \frac{x^n-a^n}{x-a} \] This isn’t quite exact - there is a gap between this linear approximation and the exact curve. However, the smaller the movement across the x-axis, the closer the approximation is to the actual curve.

For example, in the graph below, the line drawn between \(f(1)\) and \(f(2)\) is closer to the gradient at \(f(1)\) of the true curve, compared to the line drawn between \(f(1)\) and \(f(3)\).

Consequently, we want to determine the gradient at the limit i.e. where \(a \rightarrow x\), to get the true gradient.

First though, let’s factor out the \(x-a\) term for simplicity. Let’s derive a generic formula for this:

• If \(n=2\), then \(x^2-a^2 = (x-a)(x+a)\)
• If \(n=3\), then \(x^3-a^3 = (x-a)(x^2+xa+a^2)\)
• If \(n=4\), then \(x^4-a^4 = (x-a)(x^3+x^2a+xa^2+a^3)\)
• If \(n=5\), then \(x^5-a^5 = (x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)\)
• And so on. In fact for any \(n\), we can derive \(x^n-a^n = \sum_{i=1}^n \left( x^{n-i}a^{i-1} \right)\)

And we can now sub that into our formula, and the \(x-a\) cancels out:

\[ \frac{x^n-a^n}{x-a} = \frac{x-a}{x-a} \sum_{i=1}^n \left( x^{n-i}a^{i-1} \right) = \sum_{i=1}^n \left( x^{n-i}a^{i-1} \right) \]

And now let’s calculate the result in the limit, where \(x\) approaches \(a\):

\[ \lim_{x \rightarrow a} \sum_{i=1}^n \left( x^{n-i}a^{i-1} \right) \sim \sum_{i=1}^n \left( a^{n-i}a^{i-1} \right) = \sum_{i=1}^n \left( a^{n-1} \right) = na^{n-1} \]

Hence \(\frac{d}{dx} \left(x^{n} \right) = nx^{n-1}\).

Fin.